Slab design, three edges continuous, 5x4 m (Fe500)

Data and assumptions

• Clear spans: $l_x = 4$ m (short), $l_y = 5$ m (long); support width 230 mm; $f_{ck} = 20$ MPa, $f_y = 500$ MPa.
• Live load = 4 kN/m², floor finish = 1.3 kN/m², partition = 1.5 kN/m². Unit weight of concrete 25 kN/m³. Clear cover 20 mm (mild exposure, IS 456 Table 16); 10 mm bars assumed for working out $d$.
• Concrete grade is not stated; M20 is assumed.
• The imposed load of 4 kN/m² is taken as the live load; the floor finish (1.3 kN/m²) and partition load (1.5 kN/m²) are added as dead loads.
• Panel type: one long edge discontinuous; moment coefficients from IS 456 Annex D, Table 26 (case: one long edge discontinuous).

Depth and effective span

Basic span/depth ratio = 26 (IS 456 cl. 23.2.1). Trial overall depth $D = 180$ mm, so $d = 180 - 20 - 5 = 155$ mm for the short-span bars and $d_y = 155 - 10 = 145$ mm for the long-span bars.

$$l_x = \min(l_{clear} + d,\ c/c) = 4000 + 155 = 4155\ \text{mm},\quad l_y = 5000 + 155 = 5155\ \text{mm}$$ $$\frac{l_y}{l_x} = \frac{5155}{4155} = 1.241 < 2 \Rightarrow \text{two-way slab}$$

Loads (per m²)

• Self-weight = 25 × 0.180 = 4.500 kN/m²
• Total working load $w$ = 4.500 + 1.3 + 4 + 1.5 = 11.300 kN/m²
• Factored load $w_u = 1.5 \times 11.300 = 16.950$ kN/m²

Bending moments

$M = \alpha\, w_u\, l_x^2$ with $w_u l_x^2 = 16.950 \times 4.155^2 = 292.63$ kN·m/m. Coefficients are interpolated for $l_y/l_x = 1.24$.

Check of depth

$M_{u,lim} = 0.133\, f_{ck} b d^2 = 0.133 \times 20 \times 1000 \times 155^2 \times 10^{-6} = 63.91$ kN·m/m $>$ $M_{max} = 15.81$ kN·m/m. The section is under-reinforced; $d$ is adequate for flexure.

Reinforcement

$$A_{st} = \frac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \frac{4.6 M_u}{f_{ck} b d^2}}\right] b d$$

Minimum steel = 0.12% of $bD$ = 216 mm²/m. Maximum spacing = min($3d$, 300 mm) = 300 mm in both directions.

Top steel at the discontinuous edge (for the negative moment due to partial fixity, IS 456 Annex D) extends $0.1 l$ into the span. Along the discontinuous long edge (short-span direction): 50% of mid-span steel = 109 mm²/m; provide 8 mm @ 300 c/c.

Check for shear

$V_u = w_u l_x/2 = 16.950 \times 4.155/2 = 35.21$ kN/m

$\tau_v = \dfrac{V_u}{bd} = \dfrac{35214}{1000 \times 155} = 0.227$ MPa

Steel at the support: $A_{st} = 245$ mm²/m, $p_t = 0.158\%$, so $\tau_c = 0.29$ MPa (IS 456 Table 19, M20). For $D = 180$ mm, $k = 1.24$ (cl. 40.2.1.1), so $k\tau_c = 1.24 \times 0.29 = 0.36$ MPa.

$\tau_v = 0.227 < k\tau_c = 0.36$ MPa, hence the slab is safe in shear and no shear reinforcement is needed.

Check for deflection

For the short-span bars: $p_t = \dfrac{100 \times 219}{1000 \times 155} = 0.141\%$, $f_s = 0.58 f_y \dfrac{A_{st,req}}{A_{st,prov}} = 0.58 \times 500 \times \dfrac{184}{219} = 243$ N/mm².

Modification factor from IS 456 Fig. 4 (for $p_t = 0.14\%$ and $f_s = 243$ N/mm²): at such a low steel percentage the curve reaches its upper limit, so $k_t = 2.0$.

Permissible $l/d = 26 \times 2.0 = 52.0$; actual $l_x/d = 4155/155 = 26.8$. Since actual < permissible, deflection is satisfied.

Check for development length

$L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$ with $\tau_{bd} = 1.92$ MPa (M20, deformed bars, ×1.6): $L_d = \dfrac{0.87 \times 500 \times 8}{4 \times 1.92} = 453$ mm

The check $L_d \le 1.3\dfrac{M_1}{V} + L_0$ (IS 456 cl. 26.2.3.3(c)) is made at the discontinuous long edge, where the short-span bottom bars end. Alternate bars are bent up, so 109 mm²/m reach the support: $M_1 = 0.87 f_y A_{st} d\left(1 - \dfrac{A_{st} f_y}{b d f_{ck}}\right) = 7.25$ kN·m/m and $V = 35.21$ kN/m, so $1.3\dfrac{M_1}{V} = 1.3 \times \dfrac{7.25 \times 10^6}{35214} = 268$ mm.

$L_0$ = straight length beyond the centre of the support $= 230/2 - 25$ (end cover) $= 90$ mm, plus the anchorage value of any hook or bend:

• straight bars: $268 + 90 = 358$ mm $< 453$ mm (not enough);
• with a 90° bend ($8\phi = 64$ mm): $422$ mm $< 453$ mm (not enough);
• with a standard U-hook ($16\phi = 128$ mm): $268 + 90 + 128 = 486$ mm $> 453$ mm.

So the bottom bars are anchored with standard hooks at the discontinuous edge, and the development length is then satisfied. At the continuous edges the bottom bars run into the adjoining panels.

Torsion reinforcement at corners

Corner size $l_x/5 = 4155/5 = 831$ mm (say 840 mm) in both directions.

• Corners with one edge discontinuous and the other continuous (2 corners): 3/8 of the mid-span steel = 0.375 × 216 = 81 mm²/m in each layer. Provide 8 mm @ 300 c/c in each direction, top and bottom.
• No torsion steel is needed at corners where both edges are continuous (IS 456 D-1.10).

Detailing

PLAN  (long span horizontal, short span vertical)
  +==============================+
  #                              #
  # bot: 8mm@230 short span      #
  # bot: 8mm@230 long span       #
  #t                            t#
  +------------------------------+
  = / # : continuous edge (top bars)
  - / | : discontinuous edge (0.5 x mid-span top bars)
  T : torsion mesh top+bottom;  t : half mesh

SECTION ALONG SHORT SPAN
  top bars over      top bars at
  continuous edge    discontinuous edge
   ~~~~~~~            ~~~
  |------ slab D = 180 mm ------|
   bottom 8mm @ 230 c/c, alt. bars bent up,
   hooked at the discontinuous edge
  [support]            [support]

• Bottom bars: short-span bars in the lower layer, long-span bars above them; alternate bars may be bent up at $0.1 l$ from the support.
• Top bars over continuous supports extend $0.3 l_x$ from the face of the support; all bars have 20 mm cover.
• Distribution/minimum steel $\ge$ 216 mm²/m is already satisfied by the bars provided.

Answer: $D = 180$ mm; short-span bottom 8 mm @ 230 c/c; long-span bottom 8 mm @ 230 c/c; top steel over the continuous long edges 8 mm @ 205 c/c (short direction) and over the continuous short edges 8 mm @ 230 c/c (long direction); 8 mm @ 300 c/c top bars at the discontinuous edge. Shear and deflection are satisfied, and development length is satisfied with standard hooks at the discontinuous edge.